Answer:
D.
Explanation:
To solve the problem it is necessary to apply the concepts of Destructive and constructive interference. The constructive interference in tin film is given by

Where,
t = thickness
Wavelenght
m= is an integer
n= film/refractive index
We use this equaton because phase change is only present for gasoline air interface, but not at the gasoline-water interface. <em>The minimum t only would be when the value of m=0 then</em>



Therefore the correct answer is D. The minimum thickness of the film to see ab right reflection is 100nm
Answer:
0.208 N
Explanation:
We are given that


Distance,d=0.41 m
The magnitude of the net electrostatic force experienced by any charge at point 4
Net force,






Where 


Explanation:
Draw a free body diagram for each disc.
Disc A has three forces acting on it: 86.5 N up, T₁ down, and Wa down.
∑F = ma
86.5 N − T₁ − Wa = 0
Wa = 86.5 N − T₁
ma × 9.8 m/s² = 86.5 N − 55.6 N
ma = 3.2 kg
Disc B has three forces acting on it: T₁ up, T₂ down, and Wb down.
∑F = ma
T₁ − T₂ − Wb = 0
Wb = T₁ − T₂
mb × 9.8 m/s² = 55.6 N − 36.5 N
mb = 1.9 kg
Disc C has three forces acting on it: T₂ up, T₃ down, and Wc down.
∑F = ma
T₂ − T₃ − Wc = 0
Wc = T₂ − T₃
mc × 9.8 m/s² = 36.5 N − 9.6 N
mc = 2.7 kg
Disc D has two forces acting on it: T₃ up and Wd down.
∑F = ma
T₃ − Wd = 0
Wd = T₃
md × 9.8 m/s² = 9.6 N
md = 0.98 kg
well they are normally bigger than the inner planets, and they also have a bigger distance to go